Question

The experiment dipole moment of water molecule is $1.84D$ . Calculate the bond angle $H--O--H$ in water molecule , if dipole moment of $OH$ bond is $1.5D$ .

Answer 1

$\mu =\sqrt{{\mu }_1^2+{\mu }_1^2+2{\mu }_1{\mu }_{1}cos\alpha }$

In $H_{2}O$ only two dipoles equal to ${\mu }_1$ are operating due to two $O-H$ bonds .

Thus, $1.84=\sqrt{(1.5{)}^2+(1.5{)}^2+2\times \left(1.5\right)\times \left(1.5\right)cos\alpha }$

$\therefore$ $cos\alpha =-0.2476$

$\therefore$ $cos\alpha ={104}^{\circ }{20}^{\prime }$